# Titration problem?

## A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained. For a weak acid , the dissociation constant , Ka , is defined as: HA + H2O <=> ${A}^{-} + H 3 {O}^{+}$ Ka= [A-][H3O+]/[HA] The realtionship between Ka and pH is : -lgKa=pH - lg{[A-]/[HA]} and pH = -lg[H3O+] Using the information provided determine the dissociation constant ,Ka , for the weak acid HA. a) ${10}^{- 4 , 2}$ b)${10}^{- 4 , 5}$ c)${10}^{- 7 , 8}$ d)${10}^{- 11 , 5}$

Jun 21, 2018

$\text{Ka} = {10}^{- 4.5}$

#### Explanation:

The $\text{pKa}$ of a monoprotic weak acid is the same as the $\text{pH}$ halfway to its equivalence point. Here's how this shortcut works:

Sodium hydroxide $\text{NaOH}$ neutralizes the weak acid $\text{HA}$ to produce $\text{NaA}$ by the equation

"NaOH" + "HA" to "NaA" + "H"_2"O" color(grey)(" balanced.")

One formula unit of sodium hydroxide reacts with one $\text{HA}$ molecule to produce one formula unit of $\text{NaA}$. Thus

$n \left(\text{NaA")=n("HA","reacted")=n("NaOH", "added}\right)$

Appearantly
$n \left(\text{HA", "reacted")+ n("HA", "remaining") = n("HA", "initial}\right)$

Substituting $n \left(\text{HA","reacted}\right)$ in the second equation with $n \left(\text{NaA}\right)$ gives

$n \left(\text{HA","remaining") = n("HA", "initial")- n("NaA}\right)$

$n \left(\text{NaA") = n("NaOH") = n("HA", "initial}\right)$ at the equivalence point where V("NaOH", "added") = 22 color(white)(l) "ml" for this question;

Therefore halfway to that point at the titration midpoint at which $11 \textcolor{w h i t e}{l} \text{ml}$ of $\text{NaOH}$ has been added, $n \left(\text{NaA") = n("NaOH") = color(navy)(1/2)n("HA", "initial}\right)$

$n \left(\text{HA","remaining") = n("HA", "initial")- n("NaA}\right)$
color(white)(n("HA","remaining")) =color(navy)(1/2)n("HA", "initial")
color(white)(n("HA","remaining") )=n("NaA")

${\text{NaA" (aq) color(purple) to "Na"^(+) + "A}}^{-}$

$\text{NaA}$, a salt, completely disassociates as a strong electrolyte,

$n \left(\text{A"^-)=n("NaA")= n("HA}\right)$.

$\text{Ka}$ is defined as the equilibrium constant of the reaction

${\text{HA" (aq) rightleftharpoons "H"^+ (aq) + "A}}^{-} \left(a q\right)$

"Ka"="K"_"eq"=(["H"^+] color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"]))))=["H"^+]

given that $n \left(\text{A"^-)=n("NaA")= n("HA}\right)$ as previously stated.

Therefore

"Ka"=["H"^+]=10^(-"pH")=10^(-4.5)

Reference
"Titration Fundamentals", Chemistry LibreText, https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_Fundamentals

"Titration of a Weak Base with a Strong Acid", https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Base_with_a_Strong_Acid

Jun 21, 2018

${K}_{a} = {10}^{- 4.5}$

#### Explanation:

As an alternative approach, you can use the fact that at the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base.

The chemical equation that describes this neutralization reaction looks like this

${\text{HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O}}_{\left(l\right)}$

At the half-equivalence point, you add enough moles of strong base, which I've represented here by ${\text{OH}}^{-}$, to consume half of the initial number of moles of the weak acid.

Since the reaction consumes the weak acid and produces the conjugate base ${\text{A}}^{-}$ in a $1 : 1$ mole ratio, you can say that at the half-equivalence point, you have

$\left[{\text{HA"] = ["A}}^{-}\right]$

Now, notice that at the equivalence point, $\text{22 mL}$ of the strong base have been added to the weak acid solution. This tells you that it takes $\text{22 mL}$ of the strong base to consume all the moles of the weak acid present in the initial solution.

So if it takes $\text{22 mL}$ of the strong base to consume all the moles of the weak acid, it follows that if you add $\text{11 mL}$ of the strong base, you will consume exactly half of the moles of the weak acid.

This means that the half-equivalence point corresponds to $\text{11 mL}$ of the strong base added and a $\text{pH}$ of $4.5$.

The question provides you with the equation

-"lg"(K_a) = "pH" - "lg"( (["A"^(-)])/(["HA"]))

which is a version of the Henderson - Hasselbalch equation that can be used to calculate the $\text{pH}$ of a weak acid - conjugate base buffer.

Plug in the $\text{pH}$ of the solution at the half-equivalence point and the fact that $\left[{\text{HA"] = ["A}}^{-}\right]$ to get

$- \text{lg"(K_a)= 4.5 - "lg} \left(1\right)$

This gets you

$\text{lg} \left({K}_{a}\right) = - 4.5$

which implies

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{a} = {10}^{- 4.5}}}}$

See this answer for a similar question.

SIDE NOTE: The notation used here is based on the fact that

$\text{lg} \left(x\right) = {\log}_{10} \left(x\right)$